Roots of a Quadratic Equation

For this quadratic equation, the discriminant is \(D=b^2-4ac=(-2)^2-4(1)(10)=4-40=-36\). Since \(D<0\), the equation has no real roots. Option B would be correct only if \(D=0\). Exam tip: the sign of the discriminant quickly determines the nature of the roots.

A quadratic equation has equal real roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=1\), \(b=-6\), and \(c=k\), so \(D=(-6)^2-4(1)(k)=36-4k\). Setting \(D=0\) gives \(36-4k=0\), hence \(k=9\). Exam tip: for equal roots, directly apply the condition \(b^2-4ac=0\).

If the sum of the roots is \(S\) and their product is \(P\), the monic quadratic equation is \(x^2-Sx+P=0\). Here, \(S=7\) and \(P=10\), so the equation is \(x^2-7x+10=0\). Option B has the wrong sign for the sum, while options C and D interchange the sum and product. Exam tip: in a monic quadratic, the coefficient of \(x\) is the negative of the sum of the roots, and the constant term is their product.

For a quadratic equation \(ax^2+bx+c=0\), the product of the roots is \(\frac{c}{a}\). Here, \(a=1\) and \(c=40\), so the product of the two roots is \(40\). Since one root is \(5\), the other root is \(\frac{40}{5}=8\). The value 13 is the sum of the roots, not the other root. Exam tip: remember \(\alpha+\beta=-\frac{b}{a}\) and \(\alpha\beta=\frac{c}{a}\).

The sum of new roots is (2\alpha+2\beta=2(\alpha+\beta)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.

(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=36-16=20). Use identities in such questions.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the sum is (-1) and product is (-12), so the value is (\frac{1}{12}).

For a quadratic equation ax² + bx + c = 0 with roots α and β, Vieta’s relations give α + β = −b/a and αβ = c/a. In 2x² − 9x + 4 = 0, the coefficients are a = 2, b = −9, and c = 4. Therefore the sum is −(−9)/2 = 9/2, and the product is 4/2 = 2. Hence option A is correct. Option B incorrectly keeps the negative sign in the sum, option C gives the wrong sign for the product, and option D uses b and c without dividing by the leading coefficient.

If a product of factors is zero, at least one factor must be zero. The given factors are \\(x+2\\) and \\(x-5\\), so the roots are found by setting each factor equal to zero. From \\(x+2=0\\), subtracting 2 gives \\(x=-2\\). From \\(x-5=0\\), adding 5 gives \\(x=5\\). Thus the two roots are \\(-2\\) and \\(5\\).

This is the factor-zero principle: if \\((x+2)(x-5)=0\\), then either \\(x+2=0\\) or \\(x-5=0\\). The sign in a factor changes when solving for the root: \\(x+2\\) gives \\(-2\\), while \\(x-5\\) gives \\(5\\). Therefore option A is correct. The other choices result from failing to reverse the sign in one or both factors.

For a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a. Here, a = 1 and b = p, so the sum of the roots is -p. The given sum is -4 + (-4) = -8; therefore, -p = -8, giving p = 8. As a check, the product of the roots is (-4)(-4) = 16, which matches the constant term. Exam tip: Be careful with the negative sign in the formula for the sum of roots.

Putting (x=\frac{3}{2}) gives (\frac{9}{2}-\frac{15}{2}+c=0), so (c=3). For fractional roots calculate each term separately.

Substituting the coefficients gives \(x^2-10x+25=0\). Factoring it as \((x-5)^2=0\) shows that both roots are 5. Option B has the wrong sign; \((x+5)^2\) would produce a middle term of \(+10x\). Exam tip: when the discriminant \(D=b^2-4ac=0\), the two roots are equal.

If (\alpha\beta=0), then (\alpha=0) or (\beta=0). If the product is zero, always check for a zero root.

Factorise the quadratic: \(x^2-2x-15=(x-5)(x+3)\). Thus, \((x-5)(x+3)=0\) gives \(x=5\) or \(x=-3\). Therefore, the roots are 5 and -3. In option B, the signs of both roots are reversed. Exam tip: Set each linear factor equal to zero to obtain the roots.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\) and their product is \(\alpha\beta=\frac{c}{a}\). Here, \(a=1, b=-5, c=6\), so \(\alpha+\beta=5\) and \(\alpha\beta=6\). Therefore, \(\alpha+\beta+\alpha\beta=5+6=11\), making option A correct. Exam tip: Use Vieta’s formulas directly instead of solving for the roots individually.

If the sum and product of the roots are \(S\) and \(P\), respectively, the monic quadratic equation is \(x^2-Sx+P=0\). Here, \(S=0\) and \(P=-16\), so the equation is \(x^2-0x-16=0\), or \(x^2-16=0\). In option B, the product would be \(+16\), while options C and D give root sums of \(16\) and \(-16\), respectively. Exam tip: use the negative of the root sum as the coefficient of \(x\), and use the root product as the constant term.

The equation can be factorised as \(9x^2-6x+1=(3x-1)^2\). Thus, \(3x-1=0\), giving \(x=\frac{1}{3}\), which is the repeated root. Option B has the incorrect sign. Exam tip: the roots of a quadratic equation are equal when its discriminant \(D\) is zero.

Since \\(x=0\\) is a root, substituting it gives \\(a(0)^2+b(0)+c=0\\), so \\(c=0\\). Thus, a zero root requires the constant term to be zero. Option D states the opposite condition. Exam tip: whenever zero is a root of a quadratic equation, its constant term must be zero.

The sum of the roots is \(1+4=5\). A monic quadratic equation has the form \(x^2-(\text{sum of roots})x+(\text{product of roots})=0\). Therefore, the coefficient of \(x\) is \(-5\). Option B gives only the sum of the roots and misses the negative sign. Exam tip: in a monic quadratic, the coefficient of \(x\) is the negative of the sum of the roots.

The quadratic expression factors as \(2x^2+x-6=(2x-3)(x+2)\). Therefore, \((2x-3)(x+2)=0\) gives \(x=\frac{3}{2}\) or \(x=-2\), so option A is correct. In option B, the signs of both roots are reversed. As a quick exam check, the sum of the roots must be \(-\frac{b}{a}=-\frac{1}{2}\), which matches the roots in option A.